slide1
Download
Skip this Video
Download Presentation
Alkenes

Loading in 2 Seconds...

play fullscreen
1 / 61

Alkenes - PowerPoint PPT Presentation


  • 322 Views
  • Uploaded on

Alkenes. Introduction—Structure and Bonding. Alkenes are also called olefins . Alkenes contain a carbon—carbon double bond. Terminal alkenes have the double bond at the end of the carbon chain. Internal alkenes have at least one carbon atom bonded to each end of the double bond.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Alkenes' - Jims


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Alkenes

Introduction—Structure and Bonding

  • Alkenes are also called olefins.
  • Alkenes contain a carbon—carbon double bond.
  • Terminal alkenes have the double bond at the end of the carbon chain.
  • Internal alkenes have at least one carbon atom bonded to each end of the double bond.
  • Cycloalkenes contain a double bond in a ring.
slide2

Recall that the double bond consists of a  bond and a  bond.

  • Each carbon is sp2 hybridized and trigonal planar, with bond angles of approximately 120°.
slide3

Bond dissociation energies of the C—C bonds in ethane (a  bond only) and ethylene (one  and one  bond) can be used to estimate the strength of the  component of the double bond.

slide4

Cycloalkenes having fewer than eight carbon atoms have a cis geometry. A trans cycloalkene must have a carbon chain long enough to connect the ends of the double bond without introducing too much strain.

  • trans-Cyclooctene is the smallest isolable trans cycloalkene, but it is considerably less stable than cis-cyclooctene, making it one of the few alkenes having a higher energy trans isomer.
slide6

Calculating Degrees of Unsaturation

  • An acyclic alkene has the general structural formula CnH2n.
  • Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.
  • Cycloalkanes also have the general formula CnH2n.
  • Each  bond or ring removes two hydrogen atoms from a molecule, and this introduces one degree of unsaturation.
  • The number of degrees of unsaturation for a given molecular formula can be calculated by comparing the actual number of H atoms in a compound to the maximum number of H atoms possible for the number of carbons present if the molecule were a straight chain alkane.
  • This procedure gives the total number of rings and/or  bonds in a molecule.
slide8

Compounds with two double bonds are named as dienes by changing the “-ane” ending of the parent alkane to the suffix “–adiene”. Compounds with three double bonds are named as trienes, and so forth.

  • Always choose the longest chain that contains both atoms of the double bond.
  • In naming cycloalkenes, the double bond is located between C1 and C2, and the “1” is usually omitted in the name. The ring is numbered clockwise or counterclockwise to give the first substituent the lower number.
  • Compounds that contain both a double bond and a hydroxy group are named as alkenols and the chain (or ring) is numbered to give the OH group the lower number.
slide9

Figure 10.1

Naming an alkene in which

the longest carbon chain does

not contain both atoms of the

double bond

Figure 10.2

Examples of cycloalkene

nomenclature

slide11

Some alkene or alkenylsubstituents have common names.

  • The simplest alkene, CH2=CH2, named in the IUPAC system as ethene, is often called ethylene.

Figure 10.3

Naming alkenes with common

substituent names

slide12

Physical Properties

  • Most alkenes exhibit only weak van der Waals interactions, so their physical properties are similar to alkanes of comparable molecular weight.
  • Alkenes have low melting points and boiling points.
  • Melting and boiling points increase as the number of carbons increases because of increased surface area.
  • Alkenes are soluble in organic solvents and insoluble in water.
  • The C—C single bond between an alkyl group and one of the double bond carbons of an alkene is slightly polar because the sp3 hybridized alkyl carbon donates electron density to the sp2 hybridized alkenyl carbon.
slide13

A consequence of this dipole is that cis and trans isomeric alkenes often have somewhat different physical properties.

  • cis-2-Butene has a higher boiling point (4°C) than trans-2-butene (1°C).
  • In the cis isomer, the two Csp3—Csp2 bond dipoles reinforce each other, yielding a small net molecular dipole. In the trans isomer, the two bond dipoles cancel.
slide14

Preparation of Alkenes

  • Recall that alkenes can be prepared from alkyl halides and alcohols via elimination reactions.
slide15

Also recall that these elimination reactions are stereoselective and regioselective, so the most stable alkene is usually formed as the major product.

slide16

Introduction to Addition Reactions

  • The characteristic reaction of alkenes is addition—the  bond is broken and two new  bonds are formed.
  • Alkenes are electron rich, with the electron density of the  bond concentrated above and below the plane of the molecule.
  • Because alkenes are electron rich, simple alkenes do not react with nucleophiles or bases, reagents that are themselves electron rich. Alkenes react with electrophiles.
slide17

Because the carbon atoms of a double bond are both trigonal planar, the elements of X and Y can be added to them from the same side or from opposite sides.

slide18

Figure 10.8

Five addition reactions of

cyclohexene

slide19

Hydrohalogenation—Electrophilic Addition of HX

  • Two bonds are broken in this reaction—the weak  bond of the alkene and the HX bond—and two new  bonds are formed—one to H and one to X.
  • Recall that the H—X bond is polarized, with a partial positive charge on H. Because the electrophilic H end of HX is attracted to the electron-rich double bond, these reactions are called electrophilic additions.
slide21

Figure 10.9

  • Addition reactions are exothermic because the two  bonds formed in the product are stronger than the  and  bonds broken in the reactants. For example, H° for the addition of HBr to ethylene is –14 kcal/mol, as illustrated below.
slide22

The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions. In step 1, the alkene is the Lewis base that donates an electron pair to H—Br, the Lewis acid, while in step 2, Br¯ is the Lewis base that donates an electron pair to the carbocation, the Lewis acid.

slide23

Figure 10.10

  • In the representative energy diagram below, each step has its own energy barrier with a transition state energy maximum. Since step 1 has a higher energy transition state, it is rate-determining. H° for step 1 is positive because more bonds are broken than formed, whereas H° for step 2 is negative because only bond making occurs.
slide24

Hydrohalogenation—Markovnikov’s Rule

  • With an unsymmetrical alkene, HX can add to the double bond to give two constitutional isomers, but only one is actually formed:
  • This is a specific example of a general trend called Markovnikov’s rule.
  • Markovnikov’s rule states that in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom—that is, the carbon that has the greater number of H atoms to begin with.
slide25

The basis of Markovnikov’s rule is the formation of a carbocation in the rate-determining step of the mechanism.

  • In the addition of HX to an unsymmetrical alkene, the H atom is added to the less substituted carbon to form the more stable, more substituted carbocation.
slide26

According to the Hammond postulate, Path [2] is faster because formation of the carbocation is an endothermic process. Thus, the transition state to form the more stable 2°carbocation is lower in energy.

Figure 10.11

Electrophilic addition and the

Hammond postulate

slide27

Hydrohalogenation—Reaction Stereochemistry

  • Recall that trigonal planar atoms react with reagents from two directions with equal probability.
  • Achiral starting materials yield achiral products.
  • Sometimes new stereogenic centers are formed from hydrohalogenation:

A racemic mixture 

slide28

The mechanism of hydrohalogenation illustrates why two enantiomers are formed. Initial addition of H+occurs from either side of the planar double bond.

  • Both modes of addition generate the same achiralcarbocation. Either representation of this carbocation can be used to draw the second step of the mechanism.
slide29

Nucleophilic attack of Cl¯ on the trigonal planar carbocation also occurs from two different directions, forming two products, A and B, having a new stereogenic center.

  • A and B are enantiomers. Since attack from either direction occurs with equal probability, a racemic mixture of A and B is formed.
slide30

Hydrohalogenation occurs with syn and anti addition of HX.

  • The terms cis and trans refer to the arrangement of groups in a particular compound, usually an alkene or disubstitutedcycloalkene.
  • The terms syn and anti describe stereochemistry of a process—for example, how two groups are added to a double bond.
  • Addition of HX to 1,2-dimethylcyclohexene forms two new stereogenic centers, resulting in the formation of four stereoisomers (2 pairs of enantiomers).
slide31

Figure 10.12

Reaction of

1,2-dimethylcyclohexene with HCI

slide33

Hydration—Electrophilic Addition of Water

  • Hydration is the addition of water to an alkene to form an alcohol.
slide35

Alcohols add to alkenes, forming ethers by the same mechanism. For example, addition of CH3OH to 2-methylpropene, forms tert-butyl methyl ether (MTBE), a high octane fuel additive.

  • Note that there are three consequences to the formation of carbocation intermediates:
    • Markovnikov’s rule holds.
    • Addition of H and OH occurs in both syn and anti fashion.
    • Carbocation rearrangements can occur.
slide36

Halogenation—Addition of Halogen

  • Halogenation is the addition of X2 (X = Cl or Br) to an alkene to form a vicinal dihalide.
slide37

Halogens add to  bonds because halogens are polarizable.

  • The electron rich double bond induces a dipole in an approaching halogen molecule, making one halogen atom electron deficient and the other electron rich(X+—X–).
  • The electrophilic halogen atom is then attracted to the nucleophilic double bond, making addition possible.
  • Two facts demonstrate that halogenation follows a different mechanism from that of hydrohalogenation or hydration.
    • No rearrangements occur
    • Only anti addition of X2 is observed
    • These facts suggest that carbocations are not intermediates.
slide38

Carbocations are unstable because they have only six electrons around carbon. Halonium ions are unstable because of ring strain.

slide39

Halogenation—Reaction Stereochemistry

  • Consider the chlorination of cyclopentene to afford both enantiomers of trans-1,2-dichlorocyclopentane, with no cis products.
  • Initial addition of the electrophile Cl+ from (Cl2) occurs from either side of the planar double bond to form a bridged chloronium ion.
slide40

In the second step, nucleophilic attack of Cl¯ must occur from the backside.

  • Since the nucleophile attacks from below and the leaving group departs from above, the two Cl atoms in the product are oriented trans to each other.
  • Backside attack occurs with equal probability at either carbon of the three-membered ring to yield a racemic mixture.
slide41

cis-2-Butene yields two enantiomers, whereas trans-2-butene yields a single achiralmeso compound.

Figure 10.13

Halogenation of cis- and

trans-2-butene

slide42

Halohydrin Formation

Treatment of an alkene with a halogen X2 and H2O forms a halohydrin by addition of the elements of X and OH to the double bond.

slide43

Even though X¯ is formed in step [1] of the mechanism, its concentration is small compared to H2O (often the solvent), so H2O and not X¯ is the nucleophile.

slide44

Although the combination of Br2 and H2O effectively forms bromohydrins from alkenes, other reagents can also be used.

  • Bromohydrins are also formed with N-bromosuccinimide (NBS) in aqueous DMSO [(CH3)2S=O].
  • In H2O, NBS decomposes to form Br2, which then goes on to form a bromohydrin by the same reaction mechanism.
slide45

Because the bridged halonium ion is opened by backside attack of H2O, addition of X and OH occurs in an anti fashion and trans products are formed.

With unsymmetrical alkenes, the preferred product has the electrophile X+ bonded to the less substituted carbon, and the nucleophile (H2O) bonded to the more substituted carbon.

slide46

As in the acid catalyzed ring opening of epoxides, nucleophilic attack occurs at the more substituted carbon end of the bridged halonium ion because that carbon is better able to accommodate the partial positive charge in the transition state.

slide48

Hydroboration—Oxidation

Hydroboration—oxidation is a two-step reaction sequence that converts an alkene into an alcohol.

slide50

BH3 is a reactive gas that exists mostly as a dimer, diborane (B2H6). Borane is a strong Lewis acid that reacts readily with Lewis bases. For ease of handling in the laboratory, it is commonly used as a complex with tetrahydrofuran (THF).

The first step in hydroboration—oxidation is the addition of the elements of H and BH2 to the  bond of the alkene, forming an intermediate alkylborane.

slide51

The proposed mechanism involves concerted addition of H and BH2 from the same side of the planar double bond: the  bond and H—BH2 bond are broken as two new  bonds are formed.

  • Because four atoms are involved, the transition state is said to be four-centered.
slide52

Figure 10.15

Conversion of BH3

to a trialkylborane

with three

equivalents of

Because the alkylborane formed by the reaction with one equivalent of alkene still has two B—H bonds, it can react with two more equivalents of alkene to form a trialkylborane.

slide53

Since only one B—H bond is needed for hydroboration, commercially available dialkylboranes having the general structure R2BH are sometimes used instead of BH3. A common example is 9-borabicyclo[3.3.1]nonane (9-BBN).

slide55

This regioselectivity can be explained by considering steric factors. The larger boron atom bonds to the less sterically hindered, more accessible carbon atom.

  • Electronic factors are also used to explain this regioselectivity. If bond making and bond breaking are not completely symmetrical, boron bears a - charge in the transition state and carbon bears a + charge. Since alkyl groups stabilize a positive charge, the more stable transition state has the partial positive charge on the more substituted carbon.
slide56

Figure 10.16

Hydroboration of an

unsymmetrical alkene

slide57

Since alkylboranes react rapidly with water and spontaneously burn when exposed to air, they are oxidized, without isolation, with basic hydrogen peroxide (H2O2, ¯OH).

  • Oxidation replaces the C—B bond with a C—O bond, forming a new OH group with retention of configuration.
  • The overall result of this two-step sequence is syn addition of the elements of H and OH to a double bond in an “anti-Markovnikov” fashion.
slide59

Alkenes in Organic Synthesis

Suppose we wish to synthesize 1,2-dibromocyclohexane from cyclohexanol.

To solve this problem we must:

slide60

Working backwards from the product to determine the starting material from which it is made is called retrosynthetic analysis.

slide61

9.38-72

10.34-41, 46-64

ad